Word problem dont know where to begin

For an upcoming concert, each customer may purchase up to 3 child tickets and 3 adult tickets. let C be the number of child tickets purchased by a single customer. The probability distribution of the number of child tickets purchased by a single customer is given below.

b) suppose the mean and the standard deviation for the number of adult tickets purchased by a signle customer are 2 and 1.2, respectively. Assume that the number of child tickets and adult tickets purchased are independent random variables. Compute the mean and the standard deviation of the total number of adult and child tickets purchased by a single customer.

c) Suppose each child ticket costs $15 and each adult ticket costs $25. Compute the mean and the standard deviation of the total amount spent per purchase.

Anyone want to help me on how to just start this problem?I'm sure ill come back with more questions after.

nz, it obviously looks like you know your stuff. However, my math teacher most definitely, has not taught us your method yet. Nor have we learned whatever you did. I was wondering if there was a different more simpler way to do this, as really, I've never seen that before.

nz, it obviously looks like you know your stuff. However, my math teacher most definitely, has not taught us your method yet. Nor have we learned whatever you did. I was wondering if there was a different more simpler way to do this, as really, I've never seen that before.

Let me know, thanks!

Which part have you not seen before and therefore need explained in a different way?

I think i remember dong a 1-var stats with something. is that possible with this method?

Your teacher has a lot of explaining to do, then, not us. Ask your teacher what she/he has taught you so far in class can be used to do this problem. I am out of ideas if she has not yet taught you the concepts of expected value or variance.

On the other hand, surely you understand the concept of expected value? It is simply a weighted average of all the possible values a variable can take on. For example, if you have a 50% chance of earned $100 and 50% chance of earning $200, then you EXPECT to earn, on average, 0.5(100)+0.5(200) = $150. In your case, you have four possible values for your variable instead of just two - just multiple each value by its probability and add all of such products up - that's your expected.

nzmathman did forget one thing: there is another way to calculate variance. Instead of using , we can use

Using this method, variance is calculated by: where the mean

So, first figure out : = . You should get 1.

Then find the squared error of each from that mean and multiply those by their probabilities: . That is your variance.

Remember to square root that to get standard deviation. You should get for both variance and standard deviation.